第二章函數(shù)第5節(jié)冪函數(shù)與幾類特殊函數(shù)學(xué)習(xí)導(dǎo)航站核心知識(shí)庫(kù)：重難考點(diǎn)總結(jié)，梳理必背知識(shí)、歸納重點(diǎn)考點(diǎn)1冪函數(shù)★★☆☆☆考點(diǎn)2一次分式函數(shù)★★★☆☆考點(diǎn)3對(duì)勾函數(shù)y=ax+bx(a>0,b>0)考點(diǎn)4飄帶函數(shù)y=ax-bx(a>0,b>0)考點(diǎn)5高斯函數(shù)y=[x]★★★☆☆考點(diǎn)6狄利克雷函數(shù)D(x)=1,x∈Q,考點(diǎn)7最值函數(shù)的概念★★★☆☆（星級(jí)越高，重要程度越高）限時(shí)【變式訓(xùn)練】挑戰(zhàn)場(chǎng)：感知真題，檢驗(yàn)成果，考點(diǎn)追溯【知識(shí)梳理】考點(diǎn)1冪函數(shù)(1)冪函數(shù)的定義一般地,函數(shù)y=xα叫做冪函數(shù),其中x是自變量,α是常數(shù).(2)常見的五種冪函數(shù)的圖象(3)冪函數(shù)的性質(zhì)①冪函數(shù)在(0,+∞)上都有定義;②當(dāng)α>0時(shí),冪函數(shù)的圖象都過(guò)點(diǎn)(1,1)和(0,0),且在(0,+∞)上單調(diào)遞增;③當(dāng)α<0時(shí),冪函數(shù)的圖象都過(guò)點(diǎn)(1,1),且在(0,+∞)上單調(diào)遞減.考點(diǎn)2一次分式函數(shù)(1)定義:我們把形如y=cx+dax+b(a≠0(2)圖象(3)性質(zhì)①定義域:x|x≠?②對(duì)稱中心:?b③漸近線方程:x=-ba和y=c④單調(diào)性:當(dāng)ad>bc時(shí),函數(shù)在區(qū)間?∞,?ba和?ba,+∞上分別單調(diào)遞減;當(dāng)ad<考點(diǎn)3對(duì)勾函數(shù)y=ax+bx(a>0,b(1)性質(zhì)①奇偶性:奇函數(shù);②單調(diào)性:單增區(qū)間:?∞,?b單減區(qū)間:?ba,0③漸近線:y=ax和x=0.(2)圖象考點(diǎn)4飄帶函數(shù)y=ax-bx(a>0,b(1)性質(zhì)①奇偶性:奇函數(shù);②單調(diào)性:在(-∞,0),(0,+∞)上單調(diào)遞增;③漸近線:x=0.(2)圖象考點(diǎn)5高斯函數(shù)y=[x](1)定義:不超過(guò)實(shí)數(shù)x的最大整數(shù)稱為x的整數(shù)部分,記作[x],【典例】如,[3.4]=3,[-2.1]=-3,這一規(guī)定最早為數(shù)學(xué)家高斯所使用,故函數(shù)y=[x]稱為高斯函數(shù),又稱取整函數(shù).(2)性質(zhì)①定義域:R;值域:Z.②不具有單調(diào)性、奇偶性、周期性.(3)圖象考點(diǎn)6狄利克雷函數(shù)D(x)=1,x(1)定義域R;值域{0,1}.(2)奇偶性:偶函數(shù).(3)周期性:以任意正有理數(shù)為其周期,無(wú)最小正周期.(4)無(wú)法畫出函數(shù)的圖象,但其圖象客觀存在.考點(diǎn)7最值函數(shù)的概念設(shè)min{a,b}=a,a≤b,b直觀上來(lái)說(shuō)min{a,b}的作用就是求a,b的最小值,我們將其稱為最小值函數(shù),同樣,max{a,b}用來(lái)表示a,b的最大值,稱作最大值函數(shù).【名師點(diǎn)撥】1.(1)冪函數(shù)y=xα中,α的取值影響冪函數(shù)的定義域、圖象及性質(zhì);(2)冪函數(shù)的圖象一定會(huì)出現(xiàn)在第一象限內(nèi),一定不會(huì)出現(xiàn)在第四象限.2.對(duì)勾函數(shù)y=ax+bx(ab>0)極值與圖象的拐點(diǎn)可利用基本不等式求得【教材回歸】概念思考辨析+教材經(jīng)典改編1.思考辨析(在括號(hào)內(nèi)打“√”或“×”)(1)函數(shù)y=2x1(2)當(dāng)α>0時(shí),冪函數(shù)y=xα在(0,+∞)上是增函數(shù).()(3)當(dāng)n是偶數(shù)時(shí),冪函數(shù)y=xnm(m,n∈Z,且(4)函數(shù)y=x+mx的單調(diào)增區(qū)間是(-∞,-m),(m,+∞【答案】(1)×(2)√(3)√(4)×【解析】(1)由于冪函數(shù)的解析式為f(x)=xα,故y=2x13不是冪函數(shù),故(1)(4)只有m>0時(shí),y=x+mx的單調(diào)增區(qū)間才是(-∞,-m),(m,+∞2.若冪函數(shù)y=x-1,y=xm與y=xn在第一象限內(nèi)的圖象如圖所示,則m與n的取值情況為()A.-1<m<0<n<1B.-1<n<0<m<1C.-1<m<0<n<1D.-1<n<0<m<1【答案】D【解析】對(duì)于冪函數(shù)y=xα(α∈R),當(dāng)α>0時(shí),y=xα在(0,+∞)上單調(diào)遞增,且當(dāng)0<α<1時(shí),圖象上凸,所以0<m<1.當(dāng)α<0時(shí),y=xα在(0,+∞)上單調(diào)遞減.不妨令x=2,由圖象得2-1<2n,則-1<n<0.綜上可知,-1<n<0<m<1,故選D.3.(人教A必修一P91練習(xí)T2(1)改編)比較大小:(-1.5)3(-1.4)3.

【答案】<【解析】由于函數(shù)y=x3在R上單調(diào)遞增,且-1.5<-1.4.故(-1.5)3<(-1.4)3.4.設(shè)max{a,b}=a,a≥b,b,a<b,則函數(shù)f【答案】0【解析】作出f(x)的圖象如圖中所示的實(shí)線部分,由圖可知f(x)的最小值為0.【考向核心題型】考點(diǎn)一冪函數(shù)【典例】1.(2025·青島質(zhì)檢)如圖所示是函數(shù)y=xmn(m,n∈N*且互質(zhì))的圖象A.m,n是奇數(shù),且mnB.m是偶數(shù),n是奇數(shù),且mnC.m是偶數(shù),n是奇數(shù),且mnD.m是奇數(shù),n是偶數(shù),且mn【答案】B【解析】由圖象可知y=xmn且在(0,+∞)上單調(diào)遞增,結(jié)合題圖在(0,+∞)上的增長(zhǎng)趨勢(shì)可知,mn∈(0,1)且m為偶數(shù)又m,n∈N*且互質(zhì),故n是奇數(shù).故選B.【典例】2.(2025·鄭州模擬)已知a=1243,b=1323,c=1251A.a<b<c B.c<a<bC.a>b>c D.b<c<a【答案】B【解析】由a=1243,b=132得a=1423,b=132因?yàn)閮绾瘮?shù)y=x23在區(qū)間(0,+∞)上單調(diào)遞增,且15<1所以1523<1423<1323,【思維建模】1.對(duì)于冪函數(shù)圖象的掌握只要抓住在第一象限內(nèi)三條直線分第一象限為六個(gè)區(qū)域,即x=1,y=1,y=x所分區(qū)域.根據(jù)α<0,0<α<1,α=1,α>1的取值確定位置后,其余象限部分由奇偶性決定.2.在比較冪值的大小時(shí),必須結(jié)合冪值的特點(diǎn),選擇適當(dāng)?shù)暮瘮?shù),借助其單調(diào)性進(jìn)行比較.【變式訓(xùn)練】1.(2025·湖北名校聯(lián)考)已知冪函數(shù)f(x)=xm2+2m?3(m∈Z)是偶函數(shù),且f(x)在(-∞,A.-2 B.-1 C.0 D.3【答案】B【解析】因?yàn)楹瘮?shù)f(x)是偶函數(shù),且在(-∞,0)上單調(diào)遞增,所以函數(shù)f(x)在(0,+∞)上單調(diào)遞減,所以m2+2m-3<0,即(m-1)(m+3)<0,解得-3<m<1,又因?yàn)閙∈Z,所以m=-2或m=-1或m=0.當(dāng)m=0或m=-2時(shí),f(x)=x-3,此時(shí)f(x)為奇函數(shù),不滿足題意;當(dāng)m=-1時(shí),f(x)=x-4,此時(shí)f(x)為偶函數(shù),滿足題意,所以m=-1.【變式訓(xùn)練】2.已知函數(shù)f(x)的圖象如圖所示,則f(x)的解析式可能是()A.y=x12 B.yC.y=x3 D.y=x【答案】D【解析】對(duì)于A,函數(shù)y=x12=x的定義域?yàn)閇0,+∞),顯然不符合題意,故A對(duì)于B,函數(shù)y=x?12=1x的定義域?yàn)?0,+∞),顯然不符合題意,對(duì)于C,函數(shù)y=x3的定義域?yàn)镽,又y=x3為奇函數(shù),且在(0,+∞)上函數(shù)y=x3的圖象下凸遞增,故不符合題意,故C錯(cuò)誤;對(duì)于D,函數(shù)y=x13=3x又y=x13為奇函數(shù),且在(0,+∞)上函數(shù)y=x13的圖象上凸遞增,考點(diǎn)二幾類特殊函數(shù)角度1一次分式函數(shù)【典例】3.已知函數(shù)f(x)=ax+2?ax+1,其中(1)當(dāng)函數(shù)f(x)的圖象關(guān)于點(diǎn)P(-1,3)成中心對(duì)稱時(shí),求a的值;(2)若函數(shù)f(x)在(-1,+∞)上單調(diào)遞減,求a的取值范圍.【解析】(1)f(x)=ax+2?a=a+2?2a所以f(x)的對(duì)稱中心為點(diǎn)(-1,a),由題意得a=3.(2)由f(x)=ax+2?ax+1知直線x=-1為f(又由一次分式函數(shù)的性質(zhì)知,當(dāng)且僅當(dāng)1×(2-a)>1×a,即a<1時(shí),f(x)在(-1,+∞)上單調(diào)遞減,故a的取值范圍是(-∞,1).角度2對(duì)勾函數(shù)、飄帶函數(shù)【典例】4.函數(shù)f(x)=|x|-mx(m∈R【答案】C【解析】當(dāng)m=0時(shí),f(x)=|x|(x≠0),A有可能;當(dāng)m=1時(shí),f(x)=x易得f(x)在(0,+∞)上單調(diào)遞增,根據(jù)對(duì)勾函數(shù)圖象易得在(-∞,-1)上單調(diào)遞減,在(-1,0)上單調(diào)遞增,D有可能;當(dāng)m=-1時(shí),f(x)=x易得f(x)在(-∞,0)上單調(diào)遞減,在(0,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增,B有可能,所以C不可能.【典例】5.已知函數(shù)f(x)=x+ax(a∈R),方程f(x)=4在[0,+∞)有兩個(gè)解x1,x2,記g(a)=|x1-x2A.函數(shù)f(x)的值域是[0,+∞)B.若a=-1,則f(x)的增區(qū)間為[-1,0)和[1,+∞)C.若a=4,則g(a)=0D.函數(shù)g(a)的最大值為4【答案】B【解析】當(dāng)a=1時(shí),f(x)=x+f(-x)=?x?1x=x+1即f(x)為偶函數(shù),當(dāng)x>0時(shí),f(x)=x+1x則函數(shù)f(x)在(0,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增,由偶函數(shù)性質(zhì)知f(x)min=1+11=2,故A錯(cuò)誤當(dāng)a=-1時(shí),f(-x)=?x+1x=x?1x=f(x),當(dāng)x∈(0,1)時(shí),f(x)=-x+1x易知f(x)在(0,1)上單調(diào)遞減,當(dāng)x∈[1,+∞)時(shí),f(x)=x-1x易知f(x)在(1,+∞)上單調(diào)遞增,由偶函數(shù)對(duì)稱性知,f(x)的增區(qū)間為[-1,0),[1,+∞),故B正確;若a=4時(shí),f(x)=x+令f(x)=4時(shí),則x1=-2,x2=2,此時(shí)g(a)=4,故C錯(cuò)誤;若a=0時(shí),f(x)=|x|,令f(x)=4時(shí),則x=±4,g(a)=8,此時(shí)與函數(shù)g(a)的最大值為4矛盾,故D錯(cuò)誤.角度3高斯函數(shù)、狄利克雷函數(shù)、最值函數(shù)【典例】6.(多選)(2025·浙江名校聯(lián)考)高斯是德國(guó)著名的數(shù)學(xué)家,近代數(shù)學(xué)奠基者之一,享有“數(shù)學(xué)王子”的稱號(hào),用其名字命名的“高斯函數(shù)”為:設(shè)x∈R,用[x]表示不超過(guò)x的最大整數(shù),則y=[x]稱為高斯函數(shù),【典例】如[-2.1]=-3,[2.1]=2,則下列說(shuō)法正確的是()A.函數(shù)y=x-[x]在區(qū)間[k,k+1)(k∈Z)上單調(diào)遞增B.函數(shù)y=x-[x]的值域?yàn)閇0,1)C.函數(shù)y=x-[x]是R上的增函數(shù)D.x∈R,x≥[x]+1【答案】AB【解析】對(duì)于A,x∈[k,k+1),k∈Z,有[x]=k,則函數(shù)y=x-[x]=x-k在[k,k+1)上單調(diào)遞增,故A正確;對(duì)于B,C,當(dāng)x∈[0,1)時(shí),y=x,當(dāng)x∈[1,2)時(shí),y=x-1,當(dāng)x∈[2,3)時(shí),y=x-2,…故可畫出y=x-[x]的圖象如圖所示,由圖象可知B正確,C錯(cuò)誤;對(duì)于D,當(dāng)x=2時(shí),[x]+1=3,有2<[2]+1,故D不正確.【典例】7.(多選)(2025·福州質(zhì)檢)德國(guó)數(shù)學(xué)家狄利克雷在數(shù)學(xué)領(lǐng)域成就顯著,以其名字命名的函數(shù)f(x)=1,x∈Q,0,x∈?A.函數(shù)y=f(x)的圖象是兩條直線B.f(f(x))=1C.f(3)>f(1)D.?x∈R,都有f(1-x)=f(2+x)【答案】BD【解析】對(duì)于A,函數(shù)y=f(x)的圖象是斷續(xù)的點(diǎn)集,不是兩條直線,A錯(cuò)誤;對(duì)于B,當(dāng)x為有理數(shù)時(shí),f(x)=1,所以f(f(x))=f(1)=1,當(dāng)x為無(wú)理數(shù)時(shí),f(x)=0,f(f(x))=f(0)=1,B正確;對(duì)于C,f(3)=0,f(1)=1,所以f(1)>f(3),C錯(cuò)誤;對(duì)于D,由題意,函數(shù)定義域?yàn)镽,且f(-x)=f(x),所以f(x)為偶函數(shù),若x是有理數(shù),則x+T也是有理數(shù);若x是無(wú)理數(shù),則x+T也是無(wú)理數(shù);所以根據(jù)函數(shù)的表達(dá)式,任取一個(gè)不為零的有理數(shù)T,f(x+T)=f(x)對(duì)?x∈R恒成立,故f(x+2)=f(x)=f(-x)=f(1-x),所以?x∈R,都有f(1-x)=f(2+x),D正確.【典例】8.(2025·西安質(zhì)檢)已知f(x)=2x+1,g(x)=2(x+1)2,?x∈R,用M(x)表示f(x),g(x)中的較大者,記為:M(x)=max{f(x),g(x)}.當(dāng)x∈R時(shí),函數(shù)A.0 B.1 C.2 D.4【答案】B【解析】已知函數(shù)y=2x在R上單調(diào)遞增,若x+1≥(x+1)2,則-1≤x≤0;若x+1<(x+1)2,則x>0或x<-1.故當(dāng)-1≤x≤0時(shí),2x+1≥2(即f(x)≥g(x);當(dāng)x>0或x<-1時(shí),2x+1<2(即f(x)<g(x).綜上,M(x)=2當(dāng)x<-1時(shí),易知函數(shù)y=2u(u為自變量)在R上單調(diào)遞增,函數(shù)u=(x+1)2在(-∞,-1)上單調(diào)遞減,∴M(x)=2(x+1)2在(又2(?1+1)2=1,∴此時(shí)M(當(dāng)-1≤x≤0時(shí),函數(shù)y=2u(u為自變量)在R上單調(diào)遞增,函數(shù)u=x+1在[-1,0]上單調(diào)遞增,∴M(x)=2x+1在[-1,0]上單調(diào)遞增,故此時(shí)M(x)≥M(-1)=2-1+1=1.當(dāng)x>0時(shí),函數(shù)y=2u(u為自變量)在R上單調(diào)遞增,函數(shù)u=(x+1)2在(0,+∞)上單調(diào)遞增,∴M(x)=2(x+1)2在(0又2(0+1)2=2,∴此時(shí)M(綜上,M(x)的最小值為M(-1)=1.故選B.【思維建?！窟@幾類特殊的函數(shù)問(wèn)題都屬于新定義問(wèn)題,其解題思想圍繞著知識(shí)遷移,就是利用新、舊知識(shí)之間的聯(lián)系,由舊知識(shí)的思考方式領(lǐng)會(huì)新知識(shí)的思考過(guò)程,而產(chǎn)生遷移的關(guān)鍵是正確概括兩種知識(shí)之間包含的共同因素,并與函數(shù)的性質(zhì)相結(jié)合.【變式訓(xùn)練】3.函數(shù)y=11?x的圖象與函數(shù)y=2sinπx(-2≤xA.2 B.4 C.6 D.8【答案】D【解析】函數(shù)y=11?x與函數(shù)y=2sinπx(-2≤x≤4)的圖象均關(guān)于點(diǎn)(1,0)從圖象可知兩函數(shù)共有8個(gè)交點(diǎn),均關(guān)于點(diǎn)(1,0)成中心對(duì)稱,即橫坐標(biāo)之和等于8.【變式訓(xùn)練】4.設(shè)x∈R,用[x]表示不超過(guò)x的最大整數(shù),則y=[x]稱為高斯函數(shù),【典例】如:[-0.5]=-1,[1.5]=1,已知函數(shù)f(x)=4x?12-3×2x+4(0<x<2),則函數(shù)y=[A.?12,32 B.{-C.{-1,0,1,2} D.{0,1,2}【答案】B【解析】f(x)=4x?12-3×2x+令t=2x,t∈(1,4),可得g(t)=12t2-3t+=12(t-3)2-1g(t)在(1,3]上遞減,在[3,4)上遞增,當(dāng)t=3時(shí),g(t)有最小值g(3)=-12又因?yàn)間(1)=32,g(4)=0所以當(dāng)t∈(1,4)時(shí),g(t)∈?1即函數(shù)f(x)的值域?yàn)?1當(dāng)f(x)∈?12,0時(shí),[f(xf(x)∈[0,1)時(shí),[f(x)]=0;f(x)∈1,32時(shí),[f(所以y=[f(x)]的值域是{-1,0,1}.【限時(shí)訓(xùn)練】（限時(shí)：60分鐘）一、單選題1.(2025·東莞調(diào)研)若冪函數(shù)f(x)=(2m2-3m-1)xm在(0,+∞)上單調(diào)遞減,則m=()A.2 B.12C.-12 D.-【答案】C【解析】由冪函數(shù)的定義可知,2m2-3m-1=1,即2m2-3m-2=0,解得m=2或m=-12當(dāng)m=2時(shí),f(x)=x2,在(0,+∞)上單調(diào)遞增,不合題意;當(dāng)m=-12時(shí),f(x)=x?12,在(0,+∞)上單調(diào)遞減,符合題意,故m2.已知冪函數(shù)f(x)的圖象經(jīng)過(guò)點(diǎn)2,12,則函數(shù)g(x)=(x-6)f(x)在區(qū)間A.-2 B.-3 C.-4 D.-5【答案】D【解析】設(shè)冪函數(shù)f(x)=xα,α∈R,因?yàn)槠鋱D象過(guò)點(diǎn)2,1所以12=2α,解得α=-1則f(x)=x-1=1x則函數(shù)g(x)=(x-6)f(x)=x?6x=1-因?yàn)楹瘮?shù)y=-6x在12所以g(x)在12,1則當(dāng)x∈12,1時(shí),g(x)max=g(1)=3.若冪函數(shù)y=xa,y=xb,y=xc的部分圖象如圖所示,則點(diǎn)(ab-b,c2-c)所在象限是()A.第一象限 B.第二象限C.第三象限 D.第四象限【答案】C【解析】根據(jù)冪函數(shù)y=xa,y=xb,y=xc的部分圖象,得a為正偶數(shù),b為負(fù)奇數(shù),0<c<1,所以ab-b<0,c2-c<0,故點(diǎn)(ab-b,c2-c)在第三象限.4.函數(shù)y=1-1x【答案】B【解析】易知函數(shù)的對(duì)稱中心為點(diǎn)(1,1),且當(dāng)x>1時(shí),函數(shù)單調(diào)遞增,故選B.5.若函數(shù)f(x)=2x2+31+x2A.(-∞,3] B.(2,3)C.(2,3] D.[3,+∞)【答案】C【解析】f(x)=2x2+31+∵x2≥0,∴x2+1≥1,∴0<1x2+1≤1,∴f(x)∈6.(2025·哈爾濱調(diào)研)函數(shù)f(x)=(m2-m-1)·xm2+m?3是冪函數(shù),對(duì)任意x1,x2∈(0,+∞),且x1≠x2,滿足f(x1)?f(x2)x1?x2<0.若a,b∈R,且aA.恒大于0 B.恒小于0C.等于0 D.無(wú)法判斷【答案】B【解析】因?yàn)閷?duì)任意x1,x2∈(0,+∞),且x1≠x2,滿足f(x1)?f(x2)x1?x由f(x)=(m2-m-1)xm2+m?3是冪函數(shù),可得m2-m-1=1,解得m=2當(dāng)m=2時(shí),f(x)=x3在(0,+∞)上單調(diào)遞增,故不成立.當(dāng)m=-1時(shí),f(x)=x-3在(0,+∞)上單調(diào)遞減,滿足條件,故m=-1,f(x)=x-3,故f(x)為奇函數(shù).因?yàn)閍<0<b,|a|<|b|,所以0<-a<b,故f(-a)>f(b),所以-f(a)>f(b),所以f(a)+f(b)<0.故選B.7.(2025·宿州模擬)黎曼函數(shù)定義在[0,1]上,其解析式為:當(dāng)x=qp為既約真分?jǐn)?shù)(最簡(jiǎn)真分?jǐn)?shù))且p,q∈N*時(shí),R(x)=1p;當(dāng)x=0,1和(0,1)內(nèi)的無(wú)理數(shù)時(shí),R(x)=0.若函數(shù)f(x)是定義在R上的偶函數(shù),且?x∈R,f(x)+f(x+2)=0,當(dāng)x∈[0,1]時(shí),f(x)=R(x),則f(π)-fA.-25 B.-1C.15 D.【答案】C【解析】由f(x+2)=-f(x)得f(x+4)=-f(x+2),則f(x+4)=f(x),所以偶函數(shù)f(x)的周期為4,f(π)=f(π-4)=f(4-π)=R(4-π)=0,因?yàn)閒20265=f404+65=f?65=-f?65=-R45=-1所以f(π)-f20265=18.(2025·河北聯(lián)合調(diào)研)高斯函數(shù)也稱取整函數(shù),記作[x],是指不超過(guò)實(shí)數(shù)x的最大整數(shù),【典例】如[6.8]=6,[-4.1]=-5,該函數(shù)被廣泛應(yīng)用于數(shù)論、函數(shù)繪圖和計(jì)算機(jī)領(lǐng)域.若函數(shù)f(x)=log2(-x2+x+2),則當(dāng)x∈[0,1]時(shí),[f(x)]的值域?yàn)?)A.2,94 B.C.{1} D.{2}【答案】C【解析】由-x2+x+2>0,得(x+1)(x-2)<0,解得-1<x<2,則f(x)的定義域?yàn)閧x|-1<x<2},當(dāng)x∈[0,1]時(shí),令t=-x2+x+2,則t∈2,9函數(shù)y=-x2+x+2在0,12在12,1又函數(shù)u=log2t在2,94所以f(x)在0,12上單調(diào)遞增,在1所以f(x)的值域?yàn)?,log所以[f(x)]的值域?yàn)閧1}.二、多選題9.已知冪函數(shù)f(x)=xα的圖象經(jīng)過(guò)點(diǎn)(16,4),則下列說(shuō)法正確的有()A.f(x)是偶函數(shù)B.f(x)是增函數(shù)C.當(dāng)x>1時(shí),f(x)>1D.當(dāng)0<x1<x2時(shí),f(x1【答案】BCD【解析】因?yàn)閮绾瘮?shù)f(x)=xα的圖象經(jīng)過(guò)點(diǎn)(16,4),所以16α=4,則α=12,所以f(x)=x12其定義域?yàn)閇0,+∞),不關(guān)于原點(diǎn)對(duì)稱,所以該函數(shù)不是偶函數(shù),故A錯(cuò)誤;又12>0,所以f(x)是增函數(shù),故B正確當(dāng)x>1時(shí),f(x)>f(1)=1,故C正確;當(dāng)0<x1<x2時(shí),因?yàn)閒(x1fx1+x所以f(x=x1+=?x1?x2所以f(x1)+f(x210.已知狄利克雷函數(shù)D(x)=1,xA.D(x)是偶函數(shù) B.D(x)是單調(diào)函數(shù)C.D(x)的值域[0,1] D.D(π)<D(3.14)【答案】AD【解析】對(duì)于A,當(dāng)x∈Q時(shí),顯然-x∈Q,此時(shí)恒有D(x)=D(-x)=1,當(dāng)x?Q時(shí),此時(shí)x是無(wú)理數(shù),顯然-x也是無(wú)理數(shù),此時(shí)恒有D(x)=D(-x)=0,所以D(x)是偶函數(shù),因此A正確;對(duì)于B,因?yàn)镈(0)=D(1)=1,所以函數(shù)D(x)不是實(shí)數(shù)集上的單調(diào)函數(shù),因此B不正確;對(duì)于C,由函數(shù)的解析式,可知D(x)的值域?yàn)閧0,1},因此C不正確;對(duì)于D,因?yàn)镈(π)=0,D(3.14)=1,所以D(π)<D(3.14),因此D正確.11.對(duì)于函數(shù)f(x)=x1+x(x∈R)A.f(-x+1)+f(x-1)=0B.當(dāng)m∈(0,1)時(shí),方程f(x)=m有唯一實(shí)數(shù)解C.函數(shù)f(x)的值域?yàn)?-∞,+∞)D.?x1≠x2,f(【答案】ABD【解析】因?yàn)閒(-x)+f(x)=?x1+|?x+x1+x=0,令t=x-1,即f(-t)+f(t)=0,故A正確;當(dāng)x>0時(shí),f(x)=x1+x=1-所以f(x)在(0,+∞)上單調(diào)遞增,又f(0)=0,f(x)=x1+x<1,且f(x)所以f(x)的值域?yàn)?-1,1),所以f(x)的單調(diào)遞增區(qū)間為(-∞,+∞),故B正確,C錯(cuò)誤,故對(duì)?x1≠x2,f(x1)?f三、填空題12.若f(x)=x12,則不等式f(x)>f(8x-16)的解集是【答案】2,【解析】因?yàn)閒(x)=x12在定義域[0,+∞)內(nèi)為增函數(shù),且f(x)>f(8x-所以x≥0,8x?16≥0,所以不等式的解集為2,1613.函數(shù)f(x)=x2+5x2【答案】5【解析】f(x)=x2+5=x2+4+1x2+4,則t≥2,f(x)可化為y=t+1t易知該函數(shù)在[2,+∞)上單調(diào)遞增,故y=t+1t≥2+12=14.若函數(shù)f(x)=sinx+23+sinx+t(x,t∈R)的最大值記為g(t),則函數(shù)【答案】3【解析】設(shè)u=sinx+2=(sinx+3)+2sinx+3由3+sinx∈[2,4],故u∈0,3原題可化為φ(u)=|u+t|的最大值記為g(t),于是g(t)=max|=maxt,g(t)的圖象如圖所示,由y得t即g(t)的最小值為